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Heat and temperature changes. Temperature changes compared to heat energy added. Remember *this assumes NO chemical changes occur* the more heat added the more temperature change. *Unless we are at a phase change point!! The more matter present the less the temperature will change.
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Temperature changes compared to heat energy added • Remember *this assumes NO chemical changes occur* • the more heat added the more temperature change. • *Unless we are at a phase change point!! • The more matter present the less the temperature will change. • The type of matter present also has an effect on the temperature change.
Heat capacity • ~the rate of temperature change compared to the amount of heat energy added (or removed) with no chemical change for a specific substance. • Every substance absorbs heat differently. • Applying the same amount of heat to equal amounts of iron and water their rate of temperature change will differ. • (The pan gets hotter much faster than the water)
Molar heat capacity • We will mainly use molar heat capacity, which is the rate of temperature change per mole. • It’s symbol is C • It is measured in J/(K mol) • These will normally be givens in the problem.
Specific Heat Capacity • Same idea as molar heat capacity, but it is measured per gram as opposed to per mole. • It is necessary if you don’t know what the metal is you are working with. • Its symbol is c (its is written as “s” in your book, for the AP test it is c). • It is measured in J /(K g)
Throwing it all into one equation • The symbol for heat energy is q (J) • Molar heat capacity is C (J/mol K) • Temperature is T, change in temperature is T (K) • T is calculated by final temp-initial temp (Tf-Ti) • The symbol for number of particles is n (mol) • q = n C T • q = n C (Tf - Ti)
------Or------ • The symbol for heat energy is q (J) • specific heat capacity is c (J/ g K) • Temperature is T, change in temperature is T (K) • T is calculated by final temp-initial temp (Tf-Ti) • The symbol for mass m (g) • q = m c T • q = m c (Tf - Ti)
Using this equation • If 3940 J of energy is added to 43.9 mol of tungsten at 265 K, what will the final temperature be? • q = n C T • 3940 J = 43.9 mol (24.2 J/K mol) (Tf – 265 K) • Tf = 269 K • *Always make sure all units agree I will include several conversions in these problems
And the other… • 14.2 g of an unknown metal is heated with 998 J of energy. The temperature rises from 287.2 K to 366.1K, what is the metal? • q = m c T • 998 J = 14.2 g c (366.1-287.2) • c = .891 J/ g K • It is closest to aluminum on our list.
A simple one • 259 mol of sodium chloride is heated from 35o C to 68o C, how much heat was added? • q = n C T • 259mol (50.5J/molK)(341K – 308K) = q • 259mol (50.5J/molK)(33K) = q • q = 430,000 J or 430 kJ
Homework • If 3.87 kJ of heat is added to silver at 21oC and it heats to 74oC, how many moles were present?